No. For instance, take the function $\delta_n(x) = 1$ when $x \in [n, n+1]$ and $0$ otherwise.
EDIT: The following conditions will give the conclusion you want: $\delta_n(x) \geq 0$, $\int_\mathbb{R} \delta_n(x) = 1$, $\int_{-\epsilon}^{\epsilon} \delta_n(x) dx \to 1$ for all $\epsilon > 0$. I'm pretty sure other combinations of conditions will work, but I'm not sure what necessary and sufficient conditions are. That's actually an interesting question.
Here's the argument: Let
$$
E_n = \int_\mathbb{R} \phi(x) \delta_n(x) \, dx - \phi(0) = \int_\mathbb{R} (\phi(x)-\phi(0)) \delta_n(x) \, dx
$$
using the second condition. Then
$$
|E_n| \leq \int_{\epsilon}^{\epsilon} |\phi(x)-\phi(0)|\delta_n(x) \, dx + \int_{\mathbb{R} \setminus [-\epsilon, \epsilon]} |\phi(x)-\phi(0)| \delta_n(x) \, dx
$$
or
$$
|E_n| \leq \sup_{x \in [-\epsilon, \epsilon]} |\phi(x)-\phi(0)| \int_{-\epsilon}^\epsilon \delta_n(x) \, dx + 2 \sup_{x \in \mathbb{R}} |\phi(x)| \int_{\mathbb{R} \setminus [-\epsilon, \epsilon]} \delta_n(x) \, dx.
$$
By the last (and second) condition,
$$
\limsup_{n \to \infty} |E_n| \leq \sup_{x \in [-\epsilon, \epsilon]} |\phi(x)-\phi(0)|.
$$
Take $\epsilon \to 0$ and you're done.
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$$
\mbox{In spherical coordinates,}\quad
\delta\pars{\vec{r}}={\delta\pars{r}\delta\pars{\cos\pars{\theta}}\delta\pars{\phi} \over r^{2}}\quad
\mbox{such that}
$$
\begin{align}
\color{#66f}{\large\int_{{\mathbb R}^{3}}\delta\pars{\vec{r}}\,\dd^{3}\vec{r}}
&=\int_{0^{-}}^{\infty}\dd r\,r^{2}\int_{0}^{\pi}\dd\theta\,\sin\pars{\theta}
\int_{0}^{2\pi}\dd\phi\,{\delta\pars{r}\delta\pars{\cos\pars{\theta}}\delta\pars{\phi} \over r^{2}}
\\[3mm]&=\underbrace{\bracks{\int_{0^{-}}^{\infty}\delta\pars{r}\,\dd r}}
_{\ds{=\ 1}}\
\underbrace{\bracks{%
\int_{0}^{\pi}\delta\pars{\cos\pars{\theta}}\sin\pars{\theta}\,\dd\theta}}
_{\ds{=\ 1}}\
\underbrace{\bracks{\int_{0}^{2\pi}\delta\pars{\phi}\,\dd\phi}}_{\ds{=\ 1}}\
\\[3mm]&=\ \color{#66f}{\Large 1}
\end{align}
$$\mbox{Note that}\quad
\int_{{\mathbb R}^{3}}\delta\pars{\vec{r} - \vec{r}_{0}}\,\dd^{3}\vec{r}
=\int_{{\mathbb R}^{3}}\delta\pars{\vec{r}}\,\dd^{3}\vec{r}
$$
Best Answer
It is correct provided that one understand the notation $\int_{-\infty}^\infty\delta(x)\phi(x)\ dx$ correctly. This is by no means an integral and $\delta$ is not a function with real variables.
Also, one should specify in what function space is the function $\phi$ so that your identities would make sense.
Any distribution $T$ is infinitely differentiable in the sense of distribution by the definition $$ \langle \partial^\alpha T,\varphi\rangle:=(-1)^{|\alpha|}\langle \partial^\alpha\varphi,T\rangle. $$
However, it is not true that $T$ has arbitrary order weak derivative. For instance, the Heaviside step function is not weakly differentiable but its distributional derivative is the delta function.
I have never seen the word "derivable" used in that way before.(Due to change of OP)In the theory of electromagnetism, the first derivative of the delta function represents a point magnetic dipole situated at the origin. As for the higher order distributional derivatives, can I say that it is useful for demonstrating an example of distributions of which the $k$-th order derivative can be explicitly calculated?